[足式机器人]Part4 南科大高等机器人控制课 Ch03 Operator View of Rigid-Body Transformation

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B站：CLEAR_LAB
笔者带更新-运动学
课程主讲教师：
Prof. Wei Zhang

南科大高等机器人控制课 Ch03 Operator View of Rigid-Body Transformation

• [1. Rotation Operation via Differential Equation](#1. Rotation Operation via Differential Equation)
• [1.1 Skew Symmetric Matrices](#1.1 Skew Symmetric Matrices)
• • [1.2 Rotation Operation via Differential Equation](#1.2 Rotation Operation via Differential Equation)
• [1.3 Raotation Matrix as a Rotation Operator](#1.3 Raotation Matrix as a Rotation Operator)
• [2. Rotation Operation in Different Frames](#2. Rotation Operation in Different Frames)
• • [2.1 Rotation Martix Properties](#2.1 Rotation Martix Properties)
  • [2.2 Rotation Operation in Different Frames](#2.2 Rotation Operation in Different Frames)
• [3. Rigid-Body Operation via Diffeential Equation](#3. Rigid-Body Operation via Diffeential Equation)
• • [3.1 SE(3)](#3.1 SE(3))
  • [3.2 se(3)](#3.2 se(3))
• [4. Homogeneous Transformation Matrix as Rigid-Body Operator](#4. Homogeneous Transformation Matrix as Rigid-Body Operator)
• [5. Rigid-Body Operation of Screw Axis](#5. Rigid-Body Operation of Screw Axis)

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1. Rotation Operation via Differential Equation

1.1 Skew Symmetric Matrices

Recall that cross product is a special linear transformation.

For any ω ⃗ ∈ R 3 \vec{\omega}\in \mathbb{R} ^3 ω ∈R3, there is a matrix ω ⃗ ~ ∈ R 3 × 3 \tilde{\vec{\omega}}\in \mathbb{R} ^{3\times 3} ω ~∈R3×3 such that ω ⃗ × R ⃗ = ω ⃗ ~ R ⃗ \vec{\omega}\times \vec{R}=\tilde{\vec{\omega}}\vec{R} ω ×R =ω ~R
ω ⃗ = [ ω 1 ω 2 ω 3 ] ⟷ ω ⃗ ~ = [ 0 − ω 3 ω 2 ω 3 0 − ω 1 − ω 2 ω 1 0 ] \vec{\omega}=\left[ \begin{array}{c} \omega _1\\ \omega _2\\ \omega _3\\ \end{array} \right] \longleftrightarrow \tilde{\vec{\omega}}=\left[ \begin{matrix} 0& -\omega _3& \omega _2\\ \omega _3& 0& -\omega _1\\ -\omega _2& \omega _1& 0\\ \end{matrix} \right] ω = ω1ω2ω3 ⟷ω ~= 0ω3−ω2−ω30ω1ω2−ω10

Note that a ⃗ ~ = − a ⃗ ~ T \tilde{\vec{a}}=-\tilde{\vec{a}}^{\mathrm{T}} a ~=−a ~T (called skew stmmetric )
ω ⃗ ~ \tilde{\vec{\omega}} ω ~ is called a skew-symmetric matrix representation of the vector ω ⃗ \vec{\omega} ω

The set of skew-symmetric matrices in : s o ( n ) ≜ { S ∈ R n × n : S T = − S } so\left( n \right) \triangleq \left\{ S\in \mathbb{R} ^{n\times n}:S^{\mathrm{T}}=-S \right\} so(n)≜{S∈Rn×n:ST=−S}

We are interested in case n = 2 , 3 n=2,3 n=2,3

Rotation matrix ∈ S O ( 3 ) { R T R = E , det ⁡ ( R ) = 1 } \in SO\left( 3 \right) \left\{ R^{\mathrm{T}}R=E,\det \left( R \right) =1 \right\} ∈SO(3){RTR=E,det(R)=1}

1.2 Rotation Operation via Differential Equation

• Consider a point initially located at R ⃗ p 0 \vec{R}_{p_0} R p0 at time t = 0 t=0 t=0
• Rotate the point with unit angular velocity ω ^ \hat{\omega} ω^. Assuming the rotation axis passing through the origin, the motion is describe by
  R ⃗ ˙ p ( t ) = ω ^ × R ⃗ p = ω ^ ~ R ⃗ p ( t ) , p ( 0 ) = p 0 \dot{\vec{R}}_p\left( t \right) =\hat{\omega}\times \vec{R}_p=\tilde{\hat{\omega}}\vec{R}_p\left( t \right) ,p\left( 0 \right) =p_0 R ˙p(t)=ω^×R p=ω^~R p(t),p(0)=p0
  linear velocity at time t t t, recall x ˙ = A x , x ( 0 ) = x 0 ⇒ x ( t ) = e A t x 0 \dot{x}=Ax,x\left( 0 \right) =x_0\Rightarrow x\left( t \right) =e^{At}x_0 x˙=Ax,x(0)=x0⇒x(t)=eAtx0
• This is a linear ODE with solution : R ⃗ ˙ p ( t ) = ω ^ ~ R ⃗ p ( t ) ⇒ R ⃗ ˙ p ( t ) = e ω ^ ~ t R ⃗ p 0 \dot{\vec{R}}_p\left( t \right) =\tilde{\hat{\omega}}\vec{R}_p\left( t \right) \Rightarrow \dot{\vec{R}}p\left( t \right) =e^{\tilde{\hat{\omega}}t}\vec{R}{p_0} R ˙p(t)=ω^~R p(t)⇒R ˙p(t)=eω^~tR p0
• After t = θ t=\theta t=θ, the point has been rotated by θ \theta θ degree. Note R ⃗ ˙ p ( θ ) = e ω ^ ~ θ R ⃗ p 0 \dot{\vec{R}}p\left( \theta \right) =e^{\tilde{\hat{\omega}}\theta}\vec{R}{p_0} R ˙p(θ)=eω^~θR p0, e ω ^ ~ θ e^{\tilde{\hat{\omega}}\theta} eω^~θ is rotation operator
• R o t ( ω ^ , θ ) ≜ e ω ^ ~ θ \mathrm{Rot}\left( \hat{\omega},\theta \right) \triangleq e^{\tilde{\hat{\omega}}\theta} Rot(ω^,θ)≜eω^~θ can be viewed as a rotation operator that rotates a point about ω ^ \hat{\omega} ω^ through θ \theta θ degree

The discussion holds for any reference frame

1.3 Raotation Matrix as a Rotation Operator

• Theorem: Every rotation matrix R R R can be written as R = R o t ( ω ^ , θ ) ≜ e ω ^ ~ θ ∈ S O ( 3 ) { R T R = E , det ⁡ ( R ) = 1 } R=\mathrm{Rot}\left( \hat{\omega},\theta \right) \triangleq e^{\tilde{\hat{\omega}}\theta}\in SO\left( 3 \right) \left\{ R^{\mathrm{T}}R=E,\det \left( R \right) =1 \right\} R=Rot(ω^,θ)≜eω^~θ∈SO(3){RTR=E,det(R)=1}, i.e. , it represents a rotation operation about ω ^ \hat{\omega} ω^ by θ \theta θ
• We have seen how to use R R R to represent frame orientation and change of coordinate between different frames. They are quite different from the operator interpretation of R R R
• To apply the rotation operation, all the vectors / matrices have to be expressed in the same refrence frame

For example, assume R = [ 1 0 0 0 0 − 1 0 1 0 ] = R o t ( x ^ , π 2 ) R=\left[ \begin{matrix} 1& 0& 0\\ 0& 0& -1\\ 0& 1& 0\\ \end{matrix} \right] =\mathrm{Rot}\left( \hat{x},\frac{\pi}{2} \right) R= 1000010−10 =Rot(x^,2π)

Consider a relation q = R p q=Rp q=Rp:

• Change reference frame interpretation : two frames { A } , { B } \left\{ A \right\} ,\left\{ B \right\} {A},{B} , one physical Point a a a
  R R R : orientation of { B } \left\{ B \right\} {B} relative to { A } \left\{ A \right\} {A} : i.e. R = [ Q B A ] R=\left[ Q_{\mathrm{B}}^{A} \right] R=[QBA]
  then : p = a B , q = a A , q = R p ⇒ a A = [ Q B A ] a B p=a^B,q=a^A,q=Rp\Rightarrow a^A=\left[ Q_{\mathrm{B}}^{A} \right] a^B p=aB,q=aA,q=Rp⇒aA=[QBA]aB
• Rotation operator interpretation : one frame and two points
  a → a ′ , p = a A , q = a ′ A ⇒ a ′ A = R a A a\rightarrow a^{\prime},p=a^A,q={a^{\prime}}^A\Rightarrow {a^{\prime}}^A=Ra^A a→a′,p=aA,q=a′A⇒a′A=RaA

Consider the frame operation:

• Change of reference frame :
  Have one "frame object", two reference frames
  Frame object { A } \left\{ A \right\} {A}, orientation in { O } \left\{ O \right\} {O} , is R A O , R A B R_{\mathrm{A}}^{O},R_{\mathrm{A}}^{B} RAO,RAB, ⇒ R A O = R B O R A B = R R A B \Rightarrow R_{\mathrm{A}}^{O}=R_{\mathrm{B}}^{O}R_{\mathrm{A}}^{B}=RR_{\mathrm{A}}^{B} ⇒RAO=RBORAB=RRAB
• Rotation a frame : R ′ A = R R A {R^{\prime}}A=RR_A R′A=RRA
  two frame objects, one refrence frame
  more precisely : R ′ A = R R A ⇒ R A ′ O = R R A O {R^{\prime}}A=RR_A\Rightarrow R{\mathrm{A}^{\prime}}^{O}=RR{\mathrm{A}}^{O} R′A=RRA⇒RA′O=RRAO

2. Rotation Operation in Different Frames

2.1 Rotation Martix Properties

[ Q ] [ Q ] T = E \left[ Q \right] \left[ Q \right] ^{\mathrm{T}}=E [Q][Q]T=E

[ Q 1 ] [ Q 2 ] ∈ S O ( 3 ) , i f    [ Q 1 ] , [ Q 2 ] ∈ S O ( 3 ) \left[ Q_1 \right] \left[ Q_2 \right] \in SO\left( 3 \right) ,if\,\,\left[ Q_1 \right] ,\left[ Q_2 \right] \in SO\left( 3 \right) [Q1][Q2]∈SO(3),if[Q1],[Q2]∈SO(3)： product of two rotation matrices is also a rotation matrix
p , q ∈ R 3 , ∥ [ Q ] R ⃗ p − [ Q ] R ⃗ q ∥ 2 = ∥ [ Q ] ( R ⃗ p − R ⃗ q ) ∥ 2 = ( R ⃗ p − R ⃗ q ) T [ Q ] T [ Q ] ( R ⃗ p − R ⃗ q ) = ∥ R ⃗ p − R ⃗ q ∥ 2 p,q\in \mathbb{R} ^3,\left\| \left[ Q \right] \vec{R}p-\left[ Q \right] \vec{R}{\mathrm{q}} \right\| ^2=\left\| \left[ Q \right] \left( \vec{R}p-\vec{R}{\mathrm{q}} \right) \right\| ^2=\left( \vec{R}p-\vec{R}{\mathrm{q}} \right) ^{\mathrm{T}}\left[ Q \right] ^{\mathrm{T}}\left[ Q \right] \left( \vec{R}p-\vec{R}{\mathrm{q}} \right) =\left\| \vec{R}p-\vec{R}{\mathrm{q}} \right\| ^2 p,q∈R3, [Q]R p−[Q]R q 2= [Q](R p−R q) 2=(R p−R q)T[Q]T[Q](R p−R q)= R p−R q 2

∥ R ⃗ p − R ⃗ q ∥ = ∥ [ Q ] R ⃗ p − [ Q ] R ⃗ q ∥ \left\| \vec{R}p-\vec{R}{\mathrm{q}} \right\| =\left\| \left[ Q \right] \vec{R}p-\left[ Q \right] \vec{R}{\mathrm{q}} \right\| R p−R q = [Q]R p−[Q]R q : rotation operator preserves distance

[ Q ] ( v ⃗ × w ⃗ ) = [ Q ] v ⃗ × [ Q ] w ⃗ \left[ Q \right] \left( \vec{v}\times \vec{w} \right) =\left[ Q \right] \vec{v}\times \left[ Q \right] \vec{w} [Q](v ×w )=[Q]v ×[Q]w

[ Q ] w ⃗ ~ [ Q ] T = [ Q ] w ⃗ ~ \left[ Q \right] \tilde{\vec{w}}\left[ Q \right] ^{\mathrm{T}}=\widetilde{\left[ Q \right] \vec{w}} [Q]w ~[Q]T=[Q]w ------ important

2.2 Rotation Operation in Different Frames

Consider two frames { A } \left\{ A \right\} {A} and { B } \left\{ B \right\} {B}， the actual numerical values of the operator R o t ( ω ^ , θ ) \mathrm{Rot}\left( \hat{\omega},\theta \right) Rot(ω^,θ) depend on both the reference frame to repersent ω ^ \hat{\omega} ω^ and the reference frame to represent the operator itself

Consider a rotation axis ω ^ \hat{\omega} ω^ (coordinate free vector), with { A } \left\{ A \right\} {A} - frame coordinate ω ^ A \hat{\omega}^A ω^A and { B } \left\{ B \right\} {B} - frame. We know ω ⃗ A = [ Q B A ] ω ⃗ B \vec{\omega}^A=\left[ Q_{\mathrm{B}}^{A} \right] \vec{\omega}^B ω A=[QBA]ω B

Let B R o t ( B ω ^ , θ ) ^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) BRot(Bω^,θ) and A R o t ( A ω ^ , θ ) ^A\mathrm{Rot}\left( ^A\hat{\omega},\theta \right) ARot(Aω^,θ) be the two rotation matrices, representing the same rotation operatioon R o t ( ω ^ , θ ) \mathrm{Rot}\left( \hat{\omega},\theta \right) Rot(ω^,θ) in frames { A } \left\{ A \right\} {A} and { B } \left\{ B \right\} {B}

We have the relation : A R o t ( A ω ^ , θ ) = [ Q B A ] B R o t ( B ω ^ , θ ) [ Q A B ] ^A\mathrm{Rot}\left( ^A\hat{\omega},\theta \right) =\left[ Q_{\mathrm{B}}^{A} \right] ^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) \left[ Q_{\mathrm{A}}^{B} \right] ARot(Aω^,θ)=[QBA]BRot(Bω^,θ)[QAB]

• Approach 1 : two points p → p ′ ⇒ R ⃗ p ′ A = A R o t ( A ω ^ , θ ) R ⃗ p A p\rightarrow p^{\prime}\Rightarrow \vec{R}{\mathrm{p}^{\prime}}^{A}=^A\mathrm{Rot}\left( ^A\hat{\omega},\theta \right) \vec{R}{\mathrm{p}}^{A} p→p′⇒R p′A=ARot(Aω^,θ)R pA
  { B } \left\{ B \right\} {B} - frame : R ⃗ p ′ B = B R o t ( B ω ^ , θ ) R ⃗ p B \vec{R}{\mathrm{p}^{\prime}}^{B}=^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) \vec{R}{\mathrm{p}}^{B} R p′B=BRot(Bω^,θ)R pB
  ⇒ [ Q B A ] R ⃗ p ′ B = [ Q B A ] B R o t ( B ω ^ , θ ) R ⃗ p B ⇒ R ⃗ p ′ A = [ Q B A ] B R o t ( B ω ^ , θ ) [ Q B A ] R ⃗ p A ⇒ A R o t ( A ω ^ , θ ) = [ Q B A ] B R o t ( B ω ^ , θ ) [ Q B A ] \Rightarrow \left[ Q_{\mathrm{B}}^{A} \right] \vec{R}{\mathrm{p}^{\prime}}^{B}=\left[ Q{\mathrm{B}}^{A} \right] ^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) \vec{R}{\mathrm{p}}^{B}\Rightarrow \vec{R}{\mathrm{p}^{\prime}}^{A}=\left[ Q_{\mathrm{B}}^{A} \right] ^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) \left[ Q_{\mathrm{B}}^{A} \right] \vec{R}{\mathrm{p}}^{A}\Rightarrow ^A\mathrm{Rot}\left( ^A\hat{\omega},\theta \right) =\left[ Q{\mathrm{B}}^{A} \right] ^B\mathrm{Rot}\left( ^B\hat{\omega},\theta \right) \left[ Q_{\mathrm{B}}^{A} \right] ⇒[QBA]R p′B=[QBA]BRot(Bω^,θ)R pB⇒R p′A=[QBA]BRot(Bω^,θ)[QBA]R pA⇒ARot(Aω^,θ)=[QBA]BRot(Bω^,θ)[QBA]

• Approach 2 : for a ⃗ ∈ R 3 , a ⃗ ~ ∈ s o ( 3 ) , [ Q ] ∈ S O ( 3 ) \vec{a}\in \mathbb{R} ^3,\tilde{\vec{a}}\in so\left( 3 \right) ,\left[ Q \right] \in SO\left( 3 \right) a ∈R3,a ~∈so(3),[Q]∈SO(3)
  ⇒ [ Q ] a ⃗ ∈ R 3 , [ Q ] a ⃗ ~ = [ Q ] a ⃗ ~ [ Q ] T \Rightarrow \left[ Q \right] \vec{a}\in \mathbb{R} ^3,\widetilde{\left[ Q \right] \vec{a}}=\left[ Q \right] \tilde{\vec{a}}\left[ Q \right] ^{\mathrm{T}} ⇒[Q]a ∈R3,[Q]a =[Q]a ~[Q]T
  R o t ( ω ⃗ A , θ ) = e ω ⃗ ~ A θ = e [ Q B A ] ω ⃗ B ~ θ = e [ Q B A ] ω ⃗ ~ B [ Q B A ] T θ = [ Q B A ] e ω ⃗ ~ B θ [ Q B A ] T ← e P A P − 1 = P e A P − 1 Rot\left( \vec{\omega}^A,\theta \right) =e^{\tilde{\vec{\omega}}^A\theta}=e^{\widetilde{\left[ Q_{\mathrm{B}}^{A} \right] \vec{\omega}^B}\theta}=e^{\left[ Q_{\mathrm{B}}^{A} \right] \tilde{\vec{\omega}}^B\left[ Q_{\mathrm{B}}^{A} \right] ^{\mathrm{T}}\theta}=\left[ Q_{\mathrm{B}}^{A} \right] e^{\tilde{\vec{\omega}}^B\theta}\left[ Q_{\mathrm{B}}^{A} \right] ^{\mathrm{T}}\gets e^{PAP^{-1}}=Pe^AP^{-1} Rot(ω A,θ)=eω ~Aθ=e[QBA]ω B θ=e[QBA]ω ~B[QBA]Tθ=[QBA]eω ~Bθ[QBA]T←ePAP−1=PeAP−1

3. Rigid-Body Operation via Diffeential Equation

Recall: Every [ Q ] ∈ S O ( 3 ) \left[ Q \right] \in SO\left( 3 \right) [Q]∈SO(3) can be viewed as the state transition matrix associated with the rotation ODE(1). It maps the initial position to the current position(after the rotation motion)

• p ⃗ ( θ ) = R o t ( ω ⃗ , θ ) p ⃗ 0 \vec{p}\left( \theta \right) =Rot\left( \vec{\omega},\theta \right) \vec{p}_0 p (θ)=Rot(ω ,θ)p 0 viewed as solution to p ⃗ ˙ ( t ) = ω ⃗ ~ p ⃗ ( t ) , p ⃗ ( 0 ) = p ⃗ 0 \dot{\vec{p}}\left( t \right) =\tilde{\vec{\omega}}\vec{p}\left( t \right) ,\vec{p}\left( 0 \right) =\vec{p}_0 p ˙(t)=ω ~p (t),p (0)=p 0 at t = θ t=\theta t=θ
• The above relation requires that the rotation axis passes through the origin.

We can obtain similar ODE characterization for [ T ] ∈ S E ( 3 ) \left[ T \right] \in SE\left( 3 \right) [T]∈SE(3) , which will lead to exponential coordinate of SE(3)

Recall : Theorem (Chasles): Every rigid body motion can be realized by a screw motion

Consider a point p p p undergoes a screw motion with screw axis S \mathcal{S} S and unit speed ( θ ˙ = 1 \dot{\theta}=1 θ˙=1) . Let the cooresponding twist be V = θ ˙ S = ( ω ⃗ , v ⃗ ) \mathcal{V} =\dot{\theta}\mathcal{S} =\left( \vec{\omega},\vec{v} \right) V=θ˙S=(ω ,v ) . The motion can be described be the following ODE.
p ⃗ ˙ ( t ) = ω ⃗ × p ⃗ ( t ) + v ⃗ ⇒ [ p ⃗ ˙ ( t ) 0 ] = [ ω ⃗ ~ v ⃗ 0 0 ] [ p ⃗ ( t ) 1 ] \dot{\vec{p}}\left( t \right) =\vec{\omega}\times \vec{p}\left( t \right) +\vec{v}\Rightarrow \left[ \begin{array}{c} \dot{\vec{p}}\left( t \right)\\ 0\\ \end{array} \right] =\left[ \begin{matrix} \tilde{\vec{\omega}}& \vec{v}\\ 0& 0\\ \end{matrix} \right] \left[ \begin{array}{c} \vec{p}\left( t \right)\\ 1\\ \end{array} \right] p ˙(t)=ω ×p (t)+v ⇒[p ˙(t)0]=[ω ~0v 0][p (t)1]

Solution in homogeneous coordinate is :
[ p ⃗ ( t ) 1 ] = exp ⁡ ( [ ω ⃗ ~ v ⃗ 0 0 ] t ) [ p ⃗ ( 0 ) 1 ] \left[ \begin{array}{c} \vec{p}\left( t \right)\\ 1\\ \end{array} \right] =\exp \left( \left[ \begin{matrix} \tilde{\vec{\omega}}& \vec{v}\\ 0& 0\\ \end{matrix} \right] t \right) \left[ \begin{array}{c} \vec{p}\left( 0 \right)\\ 1\\ \end{array} \right] [p (t)1]=exp([ω ~0v 0]t)[p (0)1]

3.1 SE(3)

For any twist V = ( ω ⃗ , v ⃗ ) \mathcal{V} =\left( \vec{\omega},\vec{v} \right) V=(ω ,v ), let [ V ] \left[ \mathcal{V} \right] [V] be its matrix representation of twist V \mathcal{V} V
[ V ] = [ ω ⃗ ~ v ⃗ 0 0 ] ∈ R 4 × 4 \left[ \mathcal{V} \right] =\left[ \begin{matrix} \tilde{\vec{\omega}}& \vec{v}\\ 0& 0\\ \end{matrix} \right] \in \mathbb{R} ^{4\times 4} [V]=[ω ~0v 0]∈R4×4

• The abve definition also applies to a screw axis S = ( ω ⃗ , v ⃗ ) , [ S ] = [ ω ⃗ ~ v ⃗ 0 0 ] \mathcal{S} =\left( \vec{\omega},\vec{v} \right) ,\left[ \mathcal{S} \right] =\left[ \begin{matrix} \tilde{\vec{\omega}}& \vec{v}\\ 0& 0\\ \end{matrix} \right] S=(ω ,v ),[S]=[ω ~0v 0]

• With this notation, the solution is [ p ⃗ ( t ) 1 ] = e [ S ] t [ p ⃗ ( 0 ) 1 ] \left[ \begin{array}{c} \vec{p}\left( t \right)\\ 1\\ \end{array} \right] =e^{\left[ \mathcal{S} \right] t}\left[ \begin{array}{c} \vec{p}\left( 0 \right)\\ 1\\ \end{array} \right] [p (t)1]=e[S]t[p (0)1]

• Fact: e [ S ] t ∈ S E ( 3 ) e^{\left[ \mathcal{S} \right] t}\in SE\left( 3 \right) e[S]t∈SE(3) is always a valid homogeneous transformation matrix.
  [ T ] = e [ S ] t = [ [ Q ] R ⃗ 0 1 ] \left[ T \right] =e^{\left[ \mathcal{S} \right] t}=\left[ \begin{matrix} \left[ Q \right]& \vec{R}\\ 0& 1\\ \end{matrix} \right] [T]=e[S]t=[[Q]0R 1]

• Fact: Any T ∈ S E ( 3 ) T\in SE\left( 3 \right) T∈SE(3) can be written as [ T ] = e [ S ] t \left[ T \right] =e^{\left[ \mathcal{S} \right] t} [T]=e[S]t, i.e. , it can be viewed as an operator that moves a point/frame along the screw axis S \mathcal{S} S at unit speed for time t t t

3.2 se(3)

∀ ω ⃗ ∈ R 3 → ω ⃗ ~ ∈ s o ( 3 ) → e ω ⃗ ~ θ ∈ S O ( 3 ) ∀ S ∈ R 6 → [ S ] ∣ 4 × 4 = [ ω ⃗ ~ v ⃗ 0 0 ] ∈ s e ( 3 ) → e [ S ] θ ∈ S E ( 3 ) \forall \vec{\omega}\in \mathbb{R} ^3\rightarrow \tilde{\vec{\omega}}\in so\left( 3 \right) \rightarrow e^{\tilde{\vec{\omega}}\theta}\in SO\left( 3 \right) \\ \forall \mathcal{S} \in \mathbb{R} ^6\rightarrow \left. \left[ \mathcal{S} \right] \right|_{4\times 4}=\left[ \begin{matrix} \tilde{\vec{\omega}}& \vec{v}\\ 0& 0\\ \end{matrix} \right] \in se\left( 3 \right) \rightarrow e^{\left[ \mathcal{S} \right] \theta}\in SE\left( 3 \right) ∀ω ∈R3→ω ~∈so(3)→eω ~θ∈SO(3)∀S∈R6→[S]∣4×4=[ω ~0v 0]∈se(3)→e[S]θ∈SE(3)

Similar to s o ( 3 ) so\left( 3 \right) so(3) , we can define s e ( 3 ) se\left( 3 \right) se(3) :
s e ( 3 ) = { ( ω ⃗ ~ , v ⃗ ) , ω ⃗ ~ ∈ s o ( 3 ) , v ⃗ ∈ R 3 } se\left( 3 \right) =\left\{ \left( \tilde{\vec{\omega}},\vec{v} \right) ,\tilde{\vec{\omega}}\in so\left( 3 \right) ,\vec{v}\in \mathbb{R} ^3 \right\} se(3)={(ω ~,v ),ω ~∈so(3),v ∈R3}

• s e ( 3 ) se\left( 3 \right) se(3) contains all matrix representation of twists or equivalently all twists.
• In some references, [ V ] \left[ \mathcal{V} \right] [V] is called a twist
• Sometimes, we may abuse notation by writing V ∈ s e ( 3 ) \mathcal{V} \in se\left( 3 \right) V∈se(3)

4. Homogeneous Transformation Matrix as Rigid-Body Operator

• ODE for rigid motion under V = ( ω ⃗ , v ⃗ ) \mathcal{V} =\left( \vec{\omega},\vec{v} \right) V=(ω ,v )
  p ⃗ ˙ = v ⃗ + ω ⃗ × p ⃗ ⇒ [ p ⃗ ˙ 0 ] = [ ω ⃗ ~ v ⃗ 0 0 ] [ p ⃗ 1 ] ⇒ [ p ⃗ ˙ 0 ] = e [ V ] t [ p ⃗ 1 ] \dot{\vec{p}}=\vec{v}+\vec{\omega}\times \vec{p}\Rightarrow \left[ \begin{array}{c} \dot{\vec{p}}\\ 0\\ \end{array} \right] =\left[ \begin{matrix} \tilde{\vec{\omega}}& \vec{v}\\ 0& 0\\ \end{matrix} \right] \left[ \begin{array}{c} \vec{p}\\ 1\\ \end{array} \right] \Rightarrow \left[ \begin{array}{c} \dot{\vec{p}}\\ 0\\ \end{array} \right] =e^{\left[ \mathcal{V} \right] t}\left[ \begin{array}{c} \vec{p}\\ 1\\ \end{array} \right] p ˙=v +ω ×p ⇒[p ˙0]=[ω ~0v 0][p 1]⇒[p ˙0]=e[V]t[p 1]

• Consider "unit velocity" V = S \mathcal{V} =\mathcal{S} V=S, then time t t t means degree

  if not unit speed : V = θ ˙ S \mathcal{V} =\dot{\theta}\mathcal{S} V=θ˙S

• [ p ⃗ ′ 1 ] = [ T ] [ p ⃗ 1 ] \left[ \begin{array}{c} \vec{p}^{\prime}\\ 1\\ \end{array} \right] =\left[ T \right] \left[ \begin{array}{c} \vec{p}\\ 1\\ \end{array} \right] [p ′1]=[T][p 1] : "rotate" p ⃗ \vec{p} p about screw axis S \mathcal{S} S by θ \theta θ degree
  [ T ] = e [ S ] θ \left[ T \right] =e^{\left[ \mathcal{S} \right] \theta} [T]=e[S]θ, two points : [ p ⃗ 1 ] → [ p ⃗ ′ 1 ] \left[ \begin{array}{c} \vec{p}\\ 1\\ \end{array} \right] \rightarrow \left[ \begin{array}{c} \vec{p}^{\prime}\\ 1\\ \end{array} \right] [p 1]→[p ′1] , more precisely : [ p ⃗ O ′ 1 ] = [ T O ] [ p ⃗ O 1 ] \left[ \begin{array}{c} {\vec{p}^O}^{\prime}\\ 1\\ \end{array} \right] =\left[ T^O \right] \left[ \begin{array}{c} \vec{p}^O\\ 1\\ \end{array} \right] [p O′1]=[TO][p O1]

  For [ T ] ∈ S E ( 3 ) \left[ T \right] \in SE\left( 3 \right) [T]∈SE(3) config representative [ T B A ] \left[ T_{\mathrm{B}}^{A} \right] [TBA] : config of { B } \left\{ B \right\} {B} relative to { A } \left\{ A \right\} {A} ------ [ p ⃗ A 1 ] = [ T B A ] [ p ⃗ B 1 ] \left[ \begin{array}{c} \vec{p}^A\\ 1\\ \end{array} \right] =\left[ T_{\mathrm{B}}^{A} \right] \left[ \begin{array}{c} \vec{p}^B\\ 1\\ \end{array} \right] [p A1]=[TBA][p B1]------same physical point but two different frames

• [ T ] [ T A ] \left[ T \right] \left[ T_A \right] [T][TA] : "rotate" { A } \left\{ A \right\} {A}-frame about S \mathcal{S} S by θ \theta θ degree

Rigid-Body Operator in Different Frames

Expression of [ T ] \left[ T \right] [T] in another frame (other than { O } \left\{ O \right\} {O}):
[ T O ] ↔ [ T B O ] − 1 [ T O ] [ T B O ] \left[ T^O \right] \leftrightarrow \left[ T_{\mathrm{B}}^{O} \right] ^{-1}\left[ T^O \right] \left[ T_{\mathrm{B}}^{O} \right] [TO]↔[TBO]−1[TO][TBO]

5. Rigid-Body Operation of Screw Axis

Consider an arbitrary screw axis S \mathcal{S} S , suppose the axis has gone through a rigid transformation [ T ] = ( [ Q ] , R ⃗ ) \left[ T \right] =\left( \left[ Q \right] ,\vec{R} \right) [T]=([Q],R ) and the resulting new screw axis is S ′ \mathcal{S} ^{\prime} S′ , then
S ′ = [ A d T ] S \mathcal{S} ^{\prime}=\left[ Ad_T \right] \mathcal{S} S′=[AdT]S

Let's work an arbitrary frame { A } \left\{ A \right\} {A} (rigidly attached to the screw axis)

Let { B } \left\{ B \right\} {B} be the frame obtained be apply [ T ] \left[ T \right] [T] operation

the coordinate of S \mathcal{S} S in { A } \left\{ A \right\} {A} is the same as the coordinate of S ′ \mathcal{S} ^{\prime} S′ in { B } \left\{ B \right\} {B} : S A = S ′ B \mathcal{S} ^A={\mathcal{S} ^{\prime}}^B SA=S′B

We also know [ T B ] = [ T ] [ T A ] , [ T ] = [ T B A ] \left[ T_B \right] =\left[ T \right] \left[ T_{\mathrm{A}} \right] ,\left[ T \right] =\left[ T_{\mathrm{B}}^{A} \right] [TB]=[T][TA],[T]=[TBA]

Multiply [ X B A ] \left[ X_{\mathrm{B}}^{A} \right] [XBA] : ⇒ [ X B A ] S A = [ X B A ] S ′ B = S ′ A ⇒ S ′ A = [ X B A ] S A \Rightarrow \left[ X_{\mathrm{B}}^{A} \right] \mathcal{S} ^A=\left[ X_{\mathrm{B}}^{A} \right] {\mathcal{S} ^{\prime}}^B={\mathcal{S} ^{\prime}}^A\Rightarrow {\mathcal{S} ^{\prime}}^A=\left[ X_{\mathrm{B}}^{A} \right] \mathcal{S} ^A ⇒[XBA]SA=[XBA]S′B=S′A⇒S′A=[XBA]SA
[ X B A ] = [ [ Q B A ] 0 R ⃗ ~ B A [ Q B A ] [ Q B A ] ] = [ A d T ] \left[ X_{\mathrm{B}}^{A} \right] =\left[ \begin{matrix} \left[ Q_{\mathrm{B}}^{A} \right]& 0\\ \tilde{\vec{R}}{\mathrm{B}}^{A}\left[ Q{\mathrm{B}}^{A} \right]& \left[ Q_{\mathrm{B}}^{A} \right]\\ \end{matrix} \right] =\left[ Ad_T \right] [XBA]=[[QBA]R ~BA[QBA]0[QBA]]=[AdT]

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